3. Law of rational indices or intercepts (R.J. HaÜy, 1784): This law states that the intercepts made by any face of the crystal on the crystallographic axes are either (i) same as those of a unit plane (denoted by a, b, c) or (ii) some simple whole number multiples of them, e.g., la, mb and nc etc., where l, m and n are simple whole numbers, or (iii) one or two intercepts may be infinity if the face is parallel to one or two axes.
For example, in Fig.3 the intercepts made by the face LMN are 2a, 2b and 2c which are simple whole number multiples of those of unit plane.
Weiss indices : The intercepts of unit plane on X, Y and Z axes, i.e., crystal parameters are a, b and c respectively. If a face parallel to unit plane cuts the three axes with intercepts x, y and z respectively, then according to law of rational indices
x = l.a
y = m.b
and z = n.c
The coefficients (or multiples) l, m and n are called Weiss indices of the plane. The corresponding plane is described as (l m n) plane. These are generally small whole number. But this is not always true. These may be fractions of whole numbers as well as infinity.
Miller’s indices (W.H. Miller, 1839): The Miller indices of a face are the reciprocals of Weiss indices (i.e. coefficients of unit intercepts of a, b and c) expressed as integers.
Example: Suppose the intercepts made by a plane of the crystallographic axes are 3a, 3b, 2c then (i) Weiss indices are 3, 3, 2
(ii) Reciprocals of Weiss indices 1/3, 1/3, ½
(iii) Multiply reciprocals by the least common factor i.e., 6 to get small whole numbers 2, 2, 3
(iv) Hence, Miller indices are 2, 2, 3
(v) In general, Miller indices are denoted by h, k and l. Then the plane is represented by (h k l). Miller indices of crystal faces are, in general, small integers, as well as zero. Miller indices of the standard plane are always 1, 1, 1. These do not define merely, a particular plane, but a set of parallel planes. It is only the ratio of the Miller indices which is of importance, e.g., the (422) plane is the same as (211) plane; (200) plane is the same as (100) plane and so on.
The faces of a cubic crystal represented by Miller indices (100), (110) and (111) are displayed in Fig.4.
Fig. 4. Schematic description of (100), (110) and (111) planes.
X
Y
Z
(111 ) X
(110 Z
(100) Y Z
Y
X
The face BCEG or any plane parallel to it, has an intercept OG on X-axis but it is parallel to the Y- and Z- axes. If OG=1 (unity) then the Miller indices are reciprocals of (1, ∞, ∞) i.e., (1, 0, 0).
In the same manner ABCD is a (0, 1, 0) plane and CDFE is (0, 0, 1) plane. The diagonal plane AGED will have equal intercepts on OX and OY of unity, with an intercept of ∞ on the Z-axis.
Hence, this is (1, 1, 0) plane. The plane AGF makes equal intercepts with the three axes, so its Miller indices are (1, 1, 1).
Example 1. Calculate the Miller indices of crystal planes which cut through the crystal axes at (i) (2a, 3b, c) (ii) (a,b,c) (iii) (6a, 3b, 3c) (iv) (2a, -3b, -3c).
Sol. (i) The unit intercepts are a, b, and c a b c
2 3 1 intercepts (Weiss indices) 1/2 1/3 1 reciprocals
3 2 6 Miller indices
Hence Miller indices are (3 2 6) (ii)
a b c
1 1 1 intercepts 1 1 1 reciprocals
1 1 1 Miller indices
Hence Miller indices are (1 1 1) (iii)
a b c
6 3 3 intercepts 1/6 1/3 1/3 reciprocals
1 2 2 Miller indices
Hence Miller indices are (1 22) (iv)
a b c
2 -3 -3 intercepts 1/2 -1/3 -1/3 reciprocals
3 -2 -2 Miller indices
Hence Miller indices are (3 2 2 )
Example 2. Index the corresponding planes, whose intercepts on the crystallographic axes are (i) –a, -b, ∞ (ii) a/2, b/4, ∞ (iii) 2a, 3b, 4c
Sol. (i)
Intercepts -a -b ∞
Weiss indices -1 -1 ∞
Reciprocals -1 -1 0
Miller indices 1 1 0
Hence, the plane is indexed as (110) plane.
(ii)
Intercepts
2 a
4
b ∞
Weiss indices 2 1
4
1 ∞
Reciprocals 2 4 0
Miller indices 2 4 0
Hence, the plane is indexed as (2 4 0) plane.
(iii)
Intercepts 2a 3b 4c
Weiss indices 2 3 4 Reciprocals
2 1
3 1
4 1
Miller indices 6 4 3
Hence, the plane is indexed as (6 4 3) plane.
Isomorphism
E. Mitscherlich (1819) observed that a pair of salts of similar constitution, e.g., KH2PO4.H2O and KH2AsO4.H2O, in which phosphorus atom of one was replaced by an atom of arsenic in the
other, had the same crystalline form. To describe this replacement, he used the term isomorphism (Greek: same shape) and the elements were said to be isomorphous. These expressions are now applied more specifically, however, to the crystalline substances rather than to the elements contained in them. As a results of his work on the phosphates and arsenates, Mitscherlish proposed a generalization known as Mitscherlich’s law of isomorphism as “an equal number of atoms combined in the same manner produce the same crystalline form which is independent of the chemical nature of the combined atoms and determined only by their number and mode of combination”.
There are, however, many exceptions to this rule and a modified form of this law can be expressed as “the substances which are similar in crystalline form and in chemical properties can usually be represented by similar formulae”.
A better understanding of isomorphism has resulted from the investigation of crystals structure by x-ray methods. In order that two salts may be isomorphous it is not necessary that they should be chemically similar. The conditions of isomorphism are (i) the compounds must have the same formula type (ii) the respective structural units (atoms or ions) need not necessarily be of the same size but their relative sizes should be little different and should have the same polarization properties. The crystals will then have the same type of space lattices with almost identical axial ratios.
Polymorphism
Many substances exist in more than onecrystalline form. This phenomenon is known as polymorphism. The resulting shapes of the crystals depend upon the conditions of crystallization, e.g. temperature, pressure, concentration of solution, rate of cooling, etc. In case of elements, the polymorphism is called allotropy, e.g. carbon, phosphorus, sulfur, etc., exist in different allotropic form. Allotropic forms differ from one another in the number and arrangement of the structural units in the crystal lattice.
Internal arrangement of the crystals
The crystalline substances are characterized by the regular well-ordered arrangement of their constituents (ions, atoms or molecules). The arrangement can be best described in terms of space lattice, unit cell and lattice plane.
(a) Space Lattice or Crystal Lattice: A space lattice is defined as an infinite regular three dimensional array of points in which every point has surrounding identical to that of every other point. For example, a two dimensional square array of points is shown in Fig. 5. By repeated translation of two vectors a and b on the plane of paper, we can generate the square array. For this array magnitudes of a and b are equal and can be taken to be unity. These are called the fundamental translation vectors. The angle between them is 90°. Select any point in the array and look due north from this point, we will find another point at a distance of 1 unit, along north-east the nearest point at a distance of √2 units and along north north-east, the next nearest point is at a distance of √5 units. As this is true of every point in the array, the array can be called a two dimensional square lattice.
Fig 5: Two dimensional lattice where the distance between any two points is equal to na ( where n=1,2,3..)
A typical three dimensional arrangement of points i.e. space lattice is shown in Fig. 6. It is characterized by equal distance between the successive points along each of three axes. It must be noted here that a point is an imaginary and infinitesimal spot in space and hence the space lattice is an imaginary concept.
Fig 6
(b) Basis : A crystal structure is formed by associating an identical unit of assembly of particles (ions, atoms or molecules) with every space lattice point. This unit assembly is called the basis. Hence,
Crystal structure = Lattice + Basis
The planes and the inter-planar distances, dhkl which are calculated by substituting the values of h, k, and l in equations(4) for SC(simple cube), BCC(body centred cube) and FCC(face centred cube) lattices are shown in Fig. 7.
Fig. 7(i) d100 = a Fig.7(ii) d110 = a√2 Fig.7(iii) d111 = a√3
Fig. 7(iv) d200 = a/2 Fig.7(v) d220= a/2√2 Fig.7(vi) d111 = a√3
Fig. 7(vii) d200 = a/2 Fig.7(viii) d110 = a√2 Fig.7(ix)d222 = a√3
Fig. 7 (i) The shaded planes are (100) in the simple cubic lattice. d100 = a..
Fig. 7 (ii) The shaded planes are (110) in the simple cubic lattice. d110 = a√2 Fig. 7 (iii) The shaded planes are (111) in the simple cubic lattice. d111 = a√3 Fig. 7 (iv) The shaded planes are (200) in the face centred cubic lattice. d200 = a/2 Fig. 7 (v) The shaded planes are (220) in the face-centred cubic lattice. d220 = a/2√2 Fig. 7 (vi) The shaded planes are (111) in the face-centred cubic lattice. d111 = a√3 Fig. (vii) The shaded planes are (200) in the body centred cubic lattice d200 = a/2 Fig. 7 (viii) The shaded planes are (110) in the body-centred cubic lattice. d110 = a√2 Fig. 7 (ix) The shaded planes are (222) in the body-centred cubic lattice. d222 = a√3
(c) Unit cell: A unit cell is the smallest unit or geometrical portion which when repeated in space indefinitely will generate the space lattice or complete crystal. Thus, a unit cell is the fundamental building block of the lattice. For example, a unit cell is the square obtained by joining four neighbouring lattice points (Fig5). Alternatively, unit cell can be visualized with one lattice point at the centre of the square and with none at the corners (Fig. 5). In three dimensional space lattice ‘A’ box represents unit cell (Fig. 6).
The number of points per unit cell (N) is given by the expression
c f
i
N N
N= N
8 + 2 + … (2)
where Nc = points on the corners, Nf = points on the faces and Ni = point in the interior of the unit cell.
The unit cells are of the following types:
1. Simple or Primitive unit cell (P) which have the lattice points at the corners only.
2. Non-Primitive or Multiple Unit cell which have more than one lattice point. These are of the following types:
(i) Face-centred unit cell (F) which contains one point in the centre of each face besides the points at the corners of the cell.
(ii) Body-centred unit cell (I) which contains one point at the centre within its body in addition to the points at the corners of the cell.
(iii) Side-centred or End-centred unit cell (C) which contain points at the centre of any two parallel faces end faces in addition to the points at the corners of the cell.
(d) Bravais Lattices: Bravais (1848) showed that there can be only fourteen (14) different ways in which lattice points can be arranged in a three-dimensional space. These 14 arrangements are known as Bravais lattices. In a cubic system, for example, maximum three kinds of Bravais lattices are possible. These are: (i) Simple or primate cubic (SC) (ii) Face-centred cubic (FCC) and (iii) Body-centred cubic (BCC) lattices.
Calculation of number of particles in the unit cells of cubic lattices
The number of particles (ions, atoms or molecules) in the unit cell of simple, face-centred and body-centred cubic lattice are calculated as follows:
(i) In the simple unit of cell of cubic lattice: In this cell, there is one particle at each corner of the cube and each particle is equally shared between eight unit cubes. Hence the corner particle contributes only 1/8 to each cube. Hence, number of particle in one cell = 8 x 1/8 = 1
(ii) In the face-centred unit cell of cubic lattice: In this cell, there are six face-particles, one on each face and eight corner-particles, one on each corner. The fcc-centred particle is shared equally between two unit cubes and hence contributes only ½ to each cube. Hence, the total number of particles in one face-centred unit cell = (8 x 1/8) + (6 x ½) = 1 + 3 = 4
(iii) In the body-centred unit cell of cubic lattice: The body-centred particle belongs to only the single unit cell and contributes 1 to each cell. In addition to this, there are eight corner particles, each contributing 1/8 to each cell. Hence, the total number of particles in a body-centred cubic cell = (8 x 1/8) + 1 = 1 + 1 = 2
Example 3. A metallic element exists as a cubic lattice with edge length of 2.88 Å. The density of metal is 7.20 g cm-3. Calculate the number of unit cells in 100 g of the metal.
Sol. Volume of unit cell = (2.88 Å)3 = 23.888 × 10-24 cm3 Volume of 100 g of metal = 100
7.20 = 13.889 cm3 Number of unit cells in 13.889 cm3 = 13.889 cm24 3 3
23.888 10 cm× − = 5.814 × 10 23
Example 4. At room temperature, sodium crystallizes in a body-centred cubic cell with edge length, a = 4.24 Å. Calculate the theoretical density of sodium, if molar mass, Mm of sodium is equal to 23.08 mol-1.
Sol. Theoretical density, m
A
e=n M
N V
×
×
where n = number of molecules or atoms or ions in a unit cell, Mm = molar mass, NA = Avogadro’s number and V = volume of the cell.
For b.c.c. unit cell, n = 2
Volume of unit cell, V = (4.24 Å)3 = 76.225 × 10-24 cm3 Hence, p = 2 23.0823 24
6.022 10 76.225 10−
×
× × × = 1.0056 gcm-3
Example 5. A certain solid X (atomic mass 27) crystallizes in a fcc structure. The density of X is 2.70 g/cm3. Calculate the edge length ‘a’ of the unit cell of X.
Sol. Molar volume of x = 27/2.70 = 10 cm3 Volume occupied by one molecule = 10 23
6.022 10× cm3
As it is a face-centred cube (fcc), each unit cell contains 4 molecules. Hence, volume of each unit cell = 4 10 23
6.022 10
×
× cm3
If ‘a’ is the edge length, then volume of each cell = a3 Hence, a3 = 4 10 23
6.022 10
×
× = 66.423 × 10-24 cm3 or a = 4.05 × 10-8 cm = 4.05 Å
Example 6. Gold has a face-centred cubic lattice with a unit length 4.07 Å. Its density is 19.3 g/cm3. Calculate the Avogadro’s number from the data (at. wt. of Au = 197)
Sol. Number of atoms in gold fcc unit cell = 4 Let Avogadro’s number = NA
Thus, weight of one Au atom = 197/NA g Weight of one unit cell = 4 (197/NA) g
The volume of one unit cell = (4.07 x 10-8)3 cm3 Hence, density of gold cell = 4 (197/N ) g8 3A 3
(4.07 10 ) cm× − = 19.3 g/cm3 NA = 4 197 8 3
19.3 (4.07 10 )−
×
× × = 6.056 x 1023
Example 7. Nickel crystallizes in a face centred cubic crystal, the edge of the unit cell is 3.52 Å;
the atomic weight of nickel is 58.7 and its density is 8.948/cm3. Calculate Avogadro’s number.
Ans. NA= 6.02 x 1023
Example 8. An organic compound crystallizes in an orthorhombic system with two molecules per unit cell. The unit cell dimensions are 12.05, 15.05 and 2.69 Å. If the density of the crystal is 1.419 g cm-3, calculate the molar mass of the organic compound.
Sol. From equation Mm= ρ N VA n
Mm = (1.419 g cm ) (6.022 10 mol ) (12.5 15.05 2.69 10 cm )3 23 1 -24 2
− × − × × × 3
= 216.2 gmol-1
Example 9. An organic compound forms crystals of orthorhombic type with unit cell dimensions of 13.0 x 7.48 x 3.09 nm. If the density of the crystal is 1.315 x 103 kgm-3 and there are six molecules per unit cell, what is the molar mass of the compound?
Ans. 39.7 kg mol-1
Example 10 : Ag crystallizes in a cubic lattice. The density is 10.7 × 103 kg m-3. If the edge length of the unit cell is 406 pm, determine the type of the lattice.
Sol: Volume of the cell = (406 × 10-12m)3
Therefore, the lattice is f.c.c. type.
Example 11. The density of TlCl which crystallizes out in a cesium chloride lattice is 7.00 g cm
-3. What is the length of the edge of the unit cell? What is the shortest distance between Tl+ and Cl-? (Mm of TlCl = 239.82 g mol-1).
Sol. Since TlCl crystallizes out in a cesium chloride lattice which is a body-centred cubic lattice, it contains one molecule of TlCl per unit cell. Hence, volume of the unit, V =
24 3
24
(e) Lattice Planes or Net-Planes: All the points in a space lattice can be included in a set of parallel and equally spaced planes, known as lattice planes. Such a set of planes may be chosen in a large number of ways. The distance between two successive planes is represented by hkl or interplanar spacing in a crystal system is represented by the symbol, dhkl and in a cubic tetragonal and orthorhombic crystal systems, it is given by the expression
2 2 where a = unit cell edge length
The ratio of inter-planar distances of different faces in the three cubic lattices are:
Simple cube d100:d110:d111 = 1:1/2:1/3 = 1:0.707:0.577 Face centred cube d200 : d220 : d111 = 1:0.707:1.1547 Body-centred cube d200:d110:d222 = 1:414:0.577
Example 13. Calculate the interplanar spacing (dkl) for a cubic system between the following sets of planes: (a) 110 (b) 111 (c) 222. Assume that a is the edge length of the unit cell.
Example 14: The parameters of an orthorhombic unit cell are a=50 pm, b=100 pm, c=150 pm.
Determine the spacing between the (123) planes.
Sol.
or 123
123
1 3 50 pm
or d 28.87 pm
d =50 pm = 3 =
Example 15. The density of Li metal is 0.53 g cm-3 and the separation of the (100) planes of the metal is 350 pm. Determine whether the lattice is f.c.c. or b.c.c. Mm (Li) = 6.941 g mol-1
Sol. Density of Li = 0.53 g cm-3 = 530 kg m-3 For the cubic system, hkl
2 2 2 2 2 2
Example 16. The only metal that crystallizes in a primitive cubic lattice is polonium which has a unit cell side of 334.5 pm. What are the perpendicular distances between planes with Miller Indices (110), (111), (210) and (211)?
Sol. For a primitive cubic lattice 1
hkl 2 2 2 2
Example 17. Pd crystallizes in f.c.c. form. Its density is 11.9 gm/cm3. Calculate the distance between two consecutive 111 planes. Mass number of Pd = 106.4
Sol. Since Pd crystallizes in a f.c.c. form, the number of atoms of Pd in a unit cell = 4; density = 11.9 gm/cm3; Mass number of Pd = 106.4
Hence volume of a unit cell, V = a3 = 23 5.939 10 23cm3
(f) Atomic Radius (r):In simple or primitive cubic cell, the two adjacent corner atoms are supposed to touch each other. In b.c.c. cell, the atom at the centre of the cube is supposed to touch corner atoms. In f.c.c. cell, the atom at the face-centred is supposed to touch its adjacent corner atoms. The atomic radius may thus be calculated by applying simple geometry shown below:
Example 18. Copper has f.c.c. structure. Its interatomic spacing is 2.54Å. Calculate (a) the atomic radius and (b) lattice constant for Cu.
Sol. Interatomic spacing, 2r=2.54 Å Hence atomic radium, r =2.54
2 =1.27 Å
Example 19. Aluminium has a f.c.c. structure having density as 2700 Kg m-3 and atomic weight as 27. Calculate its radius.
Sol. Given P=2700 kg m-3 = 2.7 g cm-3 ; Atomic wt. Am=27; Na =6.022x10-23 mol –1 ; n=4
4.05
Example 20. Copper has f.c.c. structure with atomic radius as 1.278Å. Calculate its density (At.
Wt. of Cu = 63.5)
Example 21. Ni has a f.c.c. structure. Its density is 8.8 gm/cm3. Find out the closest distance between two Ni atoms. Mass no. of Ni = 58.71.
Sol. Since Ni has a f.c.c structure, number of Ni atoms in unit cell, n = 4; Density,l=8.8 gm/cm3;
Mass number of Ni=58.71;
Hence, volume of a unit cell,
3 The closest distance between two Ni atoms
m
(g) Packing Fraction and Empty Space in the Closest Packing: In the closest packing, spherical balls must have some vacant space in the crystal. The fraction of the total volume of the unit cell occupied by the atom(s) is known as packing fraction.
(i) In simple or primitive cubic cell, there is only one atom per unit cell.
Volume of an atom =
(ii) In body-centred cubic cell, there are two atoms per unit cell.
Volume of two atoms=
3
3 3
4 8 3a 3
2 π r π πa
3 3 4 8
⎛ ⎞
× = ⎜⎜ ⎟⎟ =
⎝ ⎠
Hence, paking fraction=
3
3
3 πa 3
8 π=0.6802=68.02%
a = 8
and vacant space in the cell=1−0.6802=0.3198=31.98%
(iii) In face-centred cubic cell, there are four atoms per unit cell.
Volume of four atoms =
3
3 3
4 16 2a 2
4 π r π πa
3 3 4 8
⎛ ⎞
× = ⎜⎜⎝ ⎟⎟⎠ =
Hence, packing fraction =
3
3
2 πa 2
6 π=0.7405=74.05%
a = 6
and vacant space in the cell = 1−0.7405=0.2595=25.95%
(h) Coordination Number (CN) : The number of nearest neighbour particles that a particle has in a crystal is called its coordination number.
(i) In a simple cubic unit cell, each particle has 6 equally spaced nearest neighbour particles.
Thus, the CN = 6
(ii) In a body-centred cubic unit cell, the particle at the centre of the cell is surrounded by 8 nearest particles situated at the corners of the cube. Thus, CN = 8
(iii) The two arrangements of closest packed layers are hexagonal close-packed structure (hcp) and cubic close-packed (ccp) or face-centred cubic structure (fcc). In each of these structures, every sphere is in contact with twelve others; six in its own layer, three in the layer above and three in the layer below. Thus, the CN = 12